What is the integral of (2x+1)^{35} ?

The integral of (2x+1)^{35} is 1/72 (2x+1)^{36}+C

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integral of (2x+1)^{35}

\int (2 x + 1)^{35} d x

\frac{1}{72} (2 x + 1)^{36} + C

Solution steps

\int (2 x + 1)^{35} d x

Apply u-substitution

= \int \frac{u ^{35}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int u^{35} d u

Apply the Power Rule

= \frac{1}{2} \cdot \frac{u ^{36}}{36}

Substitute back

u = 2 x + 1

= \frac{1}{2} \cdot \frac{( 2 x + 1 ) ^{36}}{36}

Simplify

\frac{1}{2} \cdot \frac{( 2 x + 1 ) ^{36}}{36} : \frac{1}{72} (2 x + 1)^{36}

= \frac{1}{72} (2 x + 1)^{36}

Add a constant to the solution

= \frac{1}{72} (2 x + 1)^{36} + C

What is the integral of (2x+1)^{35} ?
The integral of (2x+1)^{35} is 1/72 (2x+1)^{36}+C